SOLUTION: The perimeter of a rectangular garden is 60 m and its area is 225 m^. Find the length of the garden. Use L = length and W = width. A. Write a quadratic equation in terms of L in

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Question 1031851: The perimeter of a rectangular garden is 60 m and its area is 225 m^. Find the length of the garden. Use L = length and W = width.
A. Write a quadratic equation in terms of L in the form 0 = aL + bL + c that represents the area of the the flower bed.
B. Solve the quadratic equation you wrote in part A using one of the methods discussed in this unit.

Found 3 solutions by stanbon, mananth, MathTherapy:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
The perimeter of a rectangular garden is 60 m and its area is 225 m^. Find the length of the garden. Use L = length and W = width.
A. Write a quadratic equation in terms of L in the form 0 = aL + bL + c that represents the area of the the flower bed.
Area = L*W so 225 = L*W so W = 225/L
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P = 2(L+W)
P = 2L + 2W
0 = 2L + 2W - 60
or
0 = 2L + 225/L - 60
Quadratic:: 2L^2 - 60L + 225 = 0
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B. Solve the quadratic equation you wrote in part A using one of the methods discussed in this unit.
Comment:: Not sure what methods were discussed.
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Cheers,
Stan H.
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Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
The perimeter of a rectangular garden is 60 m and its area is 225 m^. Find the length of the garden. Use L = length and W = width.

L*W=225
L= 225/W
2(L+W) = 60
L+W=30
225/W +W =30
225 +W^2=30W
W^2-30W+225=0........................equation
L^2-30L+225=0
(W-15)^2=0
W=15
Width = 15
Length = 225/15 =15 m




Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

The perimeter of a rectangular garden is 60 m and its area is 225 m^. Find the length of the garden. Use L = length and W = width.
A. Write a quadratic equation in terms of L in the form 0 = aL + bL + c that represents the area of the the flower bed.
B. Solve the quadratic equation you wrote in part A using one of the methods discussed in this unit.
With L and W being the length and width, respectively, we get: 2(L + W) = 60______2(L + W) = 2(30)_______L + W = 30_______W = 30 - L
As area is matrix%281%2C2%2C+225%2C+m%5E2%29, we can say that: L(30 - L) = 225
30L+-+L%5E2+=+225
highlight_green%280+=+L%5E2+-+30L+%2B225%29
Solve by any of the 4 methods to get dimensions: matrix%281%2C5%2C+15%2C+m%2C+by%2C+15%2C+m%29